Given: In Trapezium ABCD, AB ∥ DC and AD = BC. Construction: Draw AD ∥ BE. Proof: In Quadrilateral ABCD, AB ∥ DE [∵ Given] AD ∥ BE [∵ By construction] Hence, ABED is a Parallelogram. AD = BE [∵ opposite sides of a parallelogram are equal] AD = BC [∵ Given] ⇒ BE = BC In ΔEBC, BE = BC [∵ Proved above]Read more
Given: In Trapezium ABCD, AB ∥ DC and AD = BC.
Construction: Draw AD ∥ BE.
Proof: In Quadrilateral ABCD,
AB ∥ DE [∵ Given]
AD ∥ BE [∵ By construction]
Hence, ABED is a Parallelogram.
AD = BE [∵ opposite sides of a parallelogram are equal]
AD = BC [∵ Given]
⇒ BE = BC
In ΔEBC, BE = BC [∵ Proved above]
Hence, ∠C = ∠2 …(1) [∵ In an isosceles triangle, the angles opposite to equal sides are equal]
∠A = ∠1 …(2) [∵ Opposite angles of a parallelogram are equal]
Here, ∠1 + ∠2 = 180° [∵ Linear pair]
⇒ ∠A + ∠C = 180° [∵ From the equation (1) and (2)]
⇒ ABED is a cyclic quadrilateral.
Construction: Join AD. Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle. Hence, ∠ADB = 90° ...(1) [∵ Angle in a semicircle is a right angle] Similarly, AC is diametar and ∠ADC is formed in semi- circle. Hence, ∠ADC = 90° ...(2) [∵ Angle in a semicircle is a right angle] Here, ∠ADB +Read more
Construction: Join AD.
Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle.
Hence, ∠ADB = 90° …(1) [∵ Angle in a semicircle is a right angle]
Similarly, AC is diametar and ∠ADC is formed in semi- circle.
Hence, ∠ADC = 90° …(2) [∵ Angle in a semicircle is a right angle]
Here, ∠ADB + ∠ADC = 90° + 90° = 180°
∠ADB and ∠ADC are forming linear pair. Therefore BDC is a straight line.
Hence, the point D lies on third side BC.
Given: Triangle ABC and ACD are two right triangle on common base AC. To prove: ∠CAD = ∠CBD. Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC. Hence, points A, B, C and D lie on the same circle. [∵ if a line segment joining two points subtends equal angles at two other points lyingRead more
Given: Triangle ABC and ACD are two right triangle on common base AC.
To prove: ∠CAD = ∠CBD.
Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC.
Hence, points A, B, C and D lie on the same circle.
[∵ if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.]
Therefore,
∠CAD = ∠CBD
[∵ Angles in the some segment are equal]
Given: Quadrilateral ABCD is a cyclic quadrilateral. To Prove: ABCD is a rectangle. Proof: In cyclic quadrilateral ABCD ∠A + ∠C = 180° ...(1) [∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.] But ∠A = ∠C ...(2) [∵ Opposite angles of a parallelogram are equal] From theRead more
Given: Quadrilateral ABCD is a cyclic quadrilateral.
To Prove: ABCD is a rectangle.
Proof: In cyclic quadrilateral ABCD
∠A + ∠C = 180° …(1)
[∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.]
But ∠A = ∠C …(2)
[∵ Opposite angles of a parallelogram are equal]
From the equation (1) and (2).
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 180°/2 = 90°
We know that, a parallelogram with one angle right angle, is a rectangle.
Hence, ABCD is a rectangle.
AC is Diameter of circle. Hence, ∠ADC = 90° and ∠ABC = 90° ...(1) [∵ Angle in a Semicircle is a right angle.] Similarly, BD is Diameter of circle. Hence, ∠BAD = 90° and ∠BCD = 90 ...(2) [∵ Angle in a semicircle is a right angle.] From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90° Hence,Read more
AC is Diameter of circle.
Hence, ∠ADC = 90° and ∠ABC = 90° …(1) [∵ Angle in a Semicircle is a right angle.]
Similarly, BD is Diameter of circle.
Hence, ∠BAD = 90° and ∠BCD = 90 …(2) [∵ Angle in a semicircle is a right angle.]
From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90°
Hence, ABCD is a rectangle.
Given : In circle C (O,r), OA = AB. To find: ∠ADB and ∠ACB. Solution: In ΔOAB OA = AB [∵ Given] OA = OB [∵ Radii of circle] Hence, OA = OB = AB ⇒ ABC is an equilateral triangle. Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°] ∠AOB = 2 ∠ADB [∵ The angle subtended by an arc at tRead more
Given : In circle C (O,r), OA = AB.
To find: ∠ADB and ∠ACB.
Solution: In ΔOAB
OA = AB [∵ Given]
OA = OB [∵ Radii of circle]
Hence, OA = OB = AB
⇒ ABC is an equilateral triangle.
Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°]
∠AOB = 2 ∠ADB
[∵ The angle subtended by an arc at the centre is double the angle subtended by it at
any]
⇒ ∠ADB = 1/2 ∠AOB ⇒ ∠ADB = (1/2) × 60° = 30
ACBD is a cyclic quadrilateral.
Therefore, ∠ACB + ∠ADB = 180°
[∵ the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠ACB + 30° = 180° [∵ ∠ADB = 30°]
⇒ ∠ACB = 180° – 30° = 150°
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B. Construction: Join PA and QA. Draw PM as bisector of chord AB. Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB. Therefore, PM ⊥AB [∵ The line drawn through the center of a circle to bisect a chord isRead more
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B.
Construction: Join PA and QA. Draw PM as bisector of chord AB.
Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the
chord.]
Hence, ∠PMA = 90°
Let, PM = X, therefore, QM = 4 – x
In ΔAPM, using Pythagoras theorem
AM² = AP² – PM² …(1)
And in ΔAPM, using Pythagoras theorem
AM² = AQ² – QM² …(2)
From the equation (1) and (2), we get
AP² – PM² = AQ² – QM²
⇒ 3² – x² = 5² -(4 – x)² ⇒ 9 – x² = 25 – (16 + x² – 8x)
⇒ 9 – 9 = 8x ⇒ x = 0/8 = 0
from the equation (1), AM² = 3² – 0² = 9 ⇒ AM = 3
⇒ AB = 2AM = 6
Given : In circle C (O, r) equal chords AB and CD intersects at P. To Prove: AP = CP and BP = DP. Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOPM and ΔONP, ∠OMP = ∠ONP [∵ Each 90] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONRead more
Given : In circle C (O, r) equal chords AB and CD intersects at P.
To Prove: AP = CP and BP = DP.
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOPM and ΔONP,
∠OMP = ∠ONP [∵ Each 90]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
PM = PN …(1) [∵ CPCT]
And AB = CD …(2) [∵ Given]
⇒ (1/2)AB = (1/2)CD
⇒ AM = CN …(3)
Adding the equations (1) and (3), we have
AM + PM = CN + PM
⇒ AP = CP …(4)
Subtracting equation (4) from (2), we have
AB – AP = CD – CP
⇒ PB = PD
Given : In circle C (O, r) equal chords AB and CD intersects at point P. To Prove: ∠OPM = ∠OPN Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOMP and ΔONP, ∠OMP = ∠ONP [∵ Each 90°] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONPRead more
Given : In circle C (O, r) equal chords AB and CD intersects at point P.
To Prove: ∠OPM = ∠OPN
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOMP and ΔONP,
∠OMP = ∠ONP [∵ Each 90°]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
∠OPM = ∠OPM [∵ CPCT]
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively. Therefore RS = SM = 6 cm Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS. Proof: In ΔORS, OS = OR and OL ⊥RS [∵ By construction] Therefore, RL = LS = 3 cm [∵ RS = 6 cm] In ΔOLS, using PythRead more
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively.
Therefore RS = SM = 6 cm
Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS.
Proof: In ΔORS,
OS = OR and OL ⊥RS [∵ By construction]
Therefore, RL = LS = 3 cm [∵ RS = 6 cm]
In ΔOLS, using Pythagoras theorem, OL² = OS² – SL²
⇒ OL² = 5² – 3² = 25 – 9 = 16
⇒ OL = 4
In ΔORK and ΔOMK,
OR = OM [∵ Radii of circle]
∠ROK = ∠MOK [∵ Equal chords subtend equal angle at the centre]
Ok = OK [∵ Common]
Hence, ∠ORK ∠OMK [∵ SAS congruency rule]
RK = MK [∵ CPCT]
Hence, OK ⊥RM
[∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]
Now, the area of ΔORS = (1/2)× RS × OL …(1)
And the area of ΔORS = (1/2) × OS × KR …(2)
From the equation (1) and (2),
1/2 × RS × OL = 1/2 × OS × KR
⇒ RS × OL = OS × KR ⇒ 6 × 4 = 5 × KR ⇒ KR = (6×4/5) = 4.8
Hence, RM = 2 × KR = 2 × 4.8 = 9.6 cm
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given: In Trapezium ABCD, AB ∥ DC and AD = BC. Construction: Draw AD ∥ BE. Proof: In Quadrilateral ABCD, AB ∥ DE [∵ Given] AD ∥ BE [∵ By construction] Hence, ABED is a Parallelogram. AD = BE [∵ opposite sides of a parallelogram are equal] AD = BC [∵ Given] ⇒ BE = BC In ΔEBC, BE = BC [∵ Proved above]Read more
Given: In Trapezium ABCD, AB ∥ DC and AD = BC.
See lessConstruction: Draw AD ∥ BE.
Proof: In Quadrilateral ABCD,
AB ∥ DE [∵ Given]
AD ∥ BE [∵ By construction]
Hence, ABED is a Parallelogram.
AD = BE [∵ opposite sides of a parallelogram are equal]
AD = BC [∵ Given]
⇒ BE = BC
In ΔEBC, BE = BC [∵ Proved above]
Hence, ∠C = ∠2 …(1) [∵ In an isosceles triangle, the angles opposite to equal sides are equal]
∠A = ∠1 …(2) [∵ Opposite angles of a parallelogram are equal]
Here, ∠1 + ∠2 = 180° [∵ Linear pair]
⇒ ∠A + ∠C = 180° [∵ From the equation (1) and (2)]
⇒ ABED is a cyclic quadrilateral.
. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Construction: Join AD. Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle. Hence, ∠ADB = 90° ...(1) [∵ Angle in a semicircle is a right angle] Similarly, AC is diametar and ∠ADC is formed in semi- circle. Hence, ∠ADC = 90° ...(2) [∵ Angle in a semicircle is a right angle] Here, ∠ADB +Read more
Construction: Join AD.
See lessProof: AB is diameter of circle and ∠ADB is Formed in semi-circle.
Hence, ∠ADB = 90° …(1) [∵ Angle in a semicircle is a right angle]
Similarly, AC is diametar and ∠ADC is formed in semi- circle.
Hence, ∠ADC = 90° …(2) [∵ Angle in a semicircle is a right angle]
Here, ∠ADB + ∠ADC = 90° + 90° = 180°
∠ADB and ∠ADC are forming linear pair. Therefore BDC is a straight line.
Hence, the point D lies on third side BC.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Given: Triangle ABC and ACD are two right triangle on common base AC. To prove: ∠CAD = ∠CBD. Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC. Hence, points A, B, C and D lie on the same circle. [∵ if a line segment joining two points subtends equal angles at two other points lyingRead more
Given: Triangle ABC and ACD are two right triangle on common base AC.
See lessTo prove: ∠CAD = ∠CBD.
Proof: Triangle ABC and ADC are on common base BC and ∠BAC = ∠BDC.
Hence, points A, B, C and D lie on the same circle.
[∵ if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.]
Therefore,
∠CAD = ∠CBD
[∵ Angles in the some segment are equal]
Prove that a cyclic parallelogram is a rectangle.
Given: Quadrilateral ABCD is a cyclic quadrilateral. To Prove: ABCD is a rectangle. Proof: In cyclic quadrilateral ABCD ∠A + ∠C = 180° ...(1) [∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.] But ∠A = ∠C ...(2) [∵ Opposite angles of a parallelogram are equal] From theRead more
Given: Quadrilateral ABCD is a cyclic quadrilateral.
See lessTo Prove: ABCD is a rectangle.
Proof: In cyclic quadrilateral ABCD
∠A + ∠C = 180° …(1)
[∵ The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.]
But ∠A = ∠C …(2)
[∵ Opposite angles of a parallelogram are equal]
From the equation (1) and (2).
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 180°/2 = 90°
We know that, a parallelogram with one angle right angle, is a rectangle.
Hence, ABCD is a rectangle.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
AC is Diameter of circle. Hence, ∠ADC = 90° and ∠ABC = 90° ...(1) [∵ Angle in a Semicircle is a right angle.] Similarly, BD is Diameter of circle. Hence, ∠BAD = 90° and ∠BCD = 90 ...(2) [∵ Angle in a semicircle is a right angle.] From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90° Hence,Read more
AC is Diameter of circle.
See lessHence, ∠ADC = 90° and ∠ABC = 90° …(1) [∵ Angle in a Semicircle is a right angle.]
Similarly, BD is Diameter of circle.
Hence, ∠BAD = 90° and ∠BCD = 90 …(2) [∵ Angle in a semicircle is a right angle.]
From the equation (1) and (2), ∠ADC = ∠ABC = ∠BAD = ∠BCD = 90°
Hence, ABCD is a rectangle.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Given : In circle C (O,r), OA = AB. To find: ∠ADB and ∠ACB. Solution: In ΔOAB OA = AB [∵ Given] OA = OB [∵ Radii of circle] Hence, OA = OB = AB ⇒ ABC is an equilateral triangle. Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°] ∠AOB = 2 ∠ADB [∵ The angle subtended by an arc at tRead more
Given : In circle C (O,r), OA = AB.
To find: ∠ADB and ∠ACB.
Solution: In ΔOAB
OA = AB [∵ Given]
OA = OB [∵ Radii of circle]
Hence, OA = OB = AB
⇒ ABC is an equilateral triangle.
Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°]
∠AOB = 2 ∠ADB
[∵ The angle subtended by an arc at the centre is double the angle subtended by it at
any]
See less⇒ ∠ADB = 1/2 ∠AOB ⇒ ∠ADB = (1/2) × 60° = 30
ACBD is a cyclic quadrilateral.
Therefore, ∠ACB + ∠ADB = 180°
[∵ the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠ACB + 30° = 180° [∵ ∠ADB = 30°]
⇒ ∠ACB = 180° – 30° = 150°
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B. Construction: Join PA and QA. Draw PM as bisector of chord AB. Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB. Therefore, PM ⊥AB [∵ The line drawn through the center of a circle to bisect a chord isRead more
Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B.
Construction: Join PA and QA. Draw PM as bisector of chord AB.
Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the
chord.]
See lessHence, ∠PMA = 90°
Let, PM = X, therefore, QM = 4 – x
In ΔAPM, using Pythagoras theorem
AM² = AP² – PM² …(1)
And in ΔAPM, using Pythagoras theorem
AM² = AQ² – QM² …(2)
From the equation (1) and (2), we get
AP² – PM² = AQ² – QM²
⇒ 3² – x² = 5² -(4 – x)² ⇒ 9 – x² = 25 – (16 + x² – 8x)
⇒ 9 – 9 = 8x ⇒ x = 0/8 = 0
from the equation (1), AM² = 3² – 0² = 9 ⇒ AM = 3
⇒ AB = 2AM = 6
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Given : In circle C (O, r) equal chords AB and CD intersects at P. To Prove: AP = CP and BP = DP. Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOPM and ΔONP, ∠OMP = ∠ONP [∵ Each 90] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONRead more
Given : In circle C (O, r) equal chords AB and CD intersects at P.
See lessTo Prove: AP = CP and BP = DP.
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOPM and ΔONP,
∠OMP = ∠ONP [∵ Each 90]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
PM = PN …(1) [∵ CPCT]
And AB = CD …(2) [∵ Given]
⇒ (1/2)AB = (1/2)CD
⇒ AM = CN …(3)
Adding the equations (1) and (3), we have
AM + PM = CN + PM
⇒ AP = CP …(4)
Subtracting equation (4) from (2), we have
AB – AP = CD – CP
⇒ PB = PD
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Given : In circle C (O, r) equal chords AB and CD intersects at point P. To Prove: ∠OPM = ∠OPN Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD. Proof: In ΔOMP and ΔONP, ∠OMP = ∠ONP [∵ Each 90°] AP = AP [∵ Common] OM = ON [∵ Equal chords of a circle are equidistant from the centre] Hence, ΔOMP ≅ ΔONPRead more
Given : In circle C (O, r) equal chords AB and CD intersects at point P.
See lessTo Prove: ∠OPM = ∠OPN
Construction: Join OP. Draw OM ⊥ AB and ON ⊥CD.
Proof: In ΔOMP and ΔONP,
∠OMP = ∠ONP [∵ Each 90°]
AP = AP [∵ Common]
OM = ON [∵ Equal chords of a circle are equidistant from the centre]
Hence, ΔOMP ≅ ΔONP [∵ RHS Congruency rule]
∠OPM = ∠OPM [∵ CPCT]
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively. Therefore RS = SM = 6 cm Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS. Proof: In ΔORS, OS = OR and OL ⊥RS [∵ By construction] Therefore, RL = LS = 3 cm [∵ RS = 6 cm] In ΔOLS, using PythRead more
Given : In figure, points R, S and M are showing the position of reshma, salma and mandeep respectively.
See lessTherefore RS = SM = 6 cm
Construction: Join OR, OS, RS, RM, and OM. Draw OL ⊥RS.
Proof: In ΔORS,
OS = OR and OL ⊥RS [∵ By construction]
Therefore, RL = LS = 3 cm [∵ RS = 6 cm]
In ΔOLS, using Pythagoras theorem, OL² = OS² – SL²
⇒ OL² = 5² – 3² = 25 – 9 = 16
⇒ OL = 4
In ΔORK and ΔOMK,
OR = OM [∵ Radii of circle]
∠ROK = ∠MOK [∵ Equal chords subtend equal angle at the centre]
Ok = OK [∵ Common]
Hence, ∠ORK ∠OMK [∵ SAS congruency rule]
RK = MK [∵ CPCT]
Hence, OK ⊥RM
[∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]
Now, the area of ΔORS = (1/2)× RS × OL …(1)
And the area of ΔORS = (1/2) × OS × KR …(2)
From the equation (1) and (2),
1/2 × RS × OL = 1/2 × OS × KR
⇒ RS × OL = OS × KR ⇒ 6 × 4 = 5 × KR ⇒ KR = (6×4/5) = 4.8
Hence, RM = 2 × KR = 2 × 4.8 = 9.6 cm